Question 472771
What is the 17th term in the arithmetic sequence in which a-sub-6 is 101 and a-sub-9 is 83? 
<pre>
{{{a[n]=a[1]+(n-1)d}}}

{{{a[6]=a[1]+(6-1)d}}}

{{{101=a[1]+5d}}}

{{{a[9]=a[1]+(9-1)d}}}

{{{83=a[1]+8d}}}

So we have this system of equations:

{{{system(101=a[1]+5d,83=a[1]+8d)}}}

Subtract the second equation from the 1st and get

{{{18=-3d}}

{{{-6=d}}}

Substitute in

{{{101=a[1]+5d}}}

{{{101=a[1]+5(-6)}}}

{{{101=a[1]-30}}}

{{{131=a[1]}}}

{{{a[n]=a[1]+(n-1)d}}}

{{{a[17]=131+(17-1)(-6)}}}

{{{a[17]=131+(16)(-6)}}}

{{{a[17]=131-96}}}

{{{a[17]=35}}}

As a check, here are the first 17 terms of the sequence:

131, 125, 119, 113, 107, <font color = "red">101</font>, 95, 89, <font color = "red">83</font>, 77, 71, 65, 59, 53, 47, 41, <font color = "red">35</font>.

The 6th, 9th and 17th terms are red:

Edwin</pre>